# Moment of Inertia Questions

1. Mar 28, 2007

### loz1588

Hi. I'm having trouble with moment of inertia in general, and so I have questions about two problems (I tried using latex, but it didn't seem to be loading properly).

The first problem:

1. The problem statement, all variables and given/known data
A slender rod with length L has a mass per unit length that varies with distance from the left-hand end, where x=0, according to dm/dx=gamma*x, where gamma has units of kg/m^2.

Calculate the total mass of the rod in terms of gamma and L.

Use I=int r^2dm to calculate the moment of inertia of the rod for an axis at the left-hand end, perpendicular to the rod. Use the expression you derived in part (a) to express I in terms of M and L.

Repeat part (b) for an axis at the right-hand end of the rod.

2. Relevant equations
I=int r^2dm
and dm/dx=gamma*x

3. The attempt at a solution
I got the first part by integrating dm/dx; the answer was (gamma*L^2)/2
I'm not sure how to do the second part. I originally solved for dm and plugged that into the integral and solved, but the program said gamma was not part of the answer. Is that correct?

The second problem:

1. The problem statement, all variables and given/known data
Find the moment of inertia of a hoop (a thin-walled, hollow ring) with mass M and radius R about an axis perpendicular to the hoop's plane at an edge.

3. The attempt at a solution
I'm not sure where to start on this one. My first issue is I'm not sure where the axis is, and once I figure out that, I'm not sure what to do.

Thanks for any help!

2. Mar 28, 2007

### Mentz114

Show your working.
In the second question, it says the axis is perpendicular to the hoop at the edge.

3. Apr 1, 2007

### loz1588

I=int r^2dm

I know that r=L/2, so I substitute that in to get I=int (L^2)/4 dm.

I then use my answer from the previous part and solve for L; I get (2M/gamma)^1/2. Plugging that into the earlier equation, I now have I=int M/(2*gamma)dm. I can pull out 1/(2*gamma) so that I have I=(1/2*gamma) int Mdm. I'm not sure of the bounds, however, but I evaluated the integral from 0 to L. My final answer is (L^2)/(4*gamma). I'm not sure if this is correct, though, because it doesn't involve the variable M.

Once I have the correct answer, though, is it true that the moment of inertia for an axis at the right-hand end of the rod is simply the negative of that answer?

4. Apr 2, 2007

### Mentz114

Check your units. Remember that gamma is mass/ unit volume.

I don't think so. Try holding a baseball bat at the wrong end. It's a lot easier to swing meaning the MoI is a lot less.

Your problem can probably be solved by working out the center of gravity of the rod, and then assuming all the mass is there.

5. Apr 3, 2007

### loz1588

I'm not quite sure what you mean by check your units. Are you saying plug in M/V for gamma? But then won't I have V in the final answer, which isn't allowed?

6. Apr 3, 2007

### Mentz114

First, I see from the question that gamma has units mass/area, not mass/volume so the units do add up.

I agree with your calculations for the first question but it doesn't make sense to me. If the centre of mass was exactly in the middle of the rod, then the MoI will be be tha same at both ends. But if it's say, 1/3 along from end, then the MoI from one end is M(L/3)^2 and from the other it's M(2L/3)^2.

I don't see how this ties in with

$$I = \int_0^L \gamma x^2 dx$$

unless the limits are changed to reflect the centre of mass.

I'm sorry I thought this was going to be easy and I hope I haven't confused the issue. Let's hope someone else can help.

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?